I would like to know how far field traveling electromagnetic radio waves separate from the near field non-traveling electromagnetic fields at a radio transmitter antenna. How is a non-traveling electromagnetic field converted into a traveling electromagnetic field?

  • $\begingroup$ Even the near field has "travelling" components as you put it. $\endgroup$ – Raziman T V Jan 15 '17 at 15:45
  • $\begingroup$ Yes, but they only travel to very restricted locations near the source antenna. $\endgroup$ – John Petrovic Jan 15 '17 at 15:48
  • $\begingroup$ I'm having a hard time getting a decent answer to this simple question from the usual sources on the internet. Perhaps physics has no good answer at the present time. That's why I have asked the question on this physics forum. $\endgroup$ – John Petrovic Jan 15 '17 at 15:53
  • $\begingroup$ or perhaps it is the wrong question. (well that's what always happens to me). Is they answer they separate because they don;t interact anyway and because they travel away from each other? If not why not. $\endgroup$ – JMLCarter Jan 15 '17 at 15:58

You are confused. The near-field and far-field components are both present at all distances from the antenna. The difference is that whilst the near-field component(s) fall of in amplitude as $\sim r^{-2}$, the far-field components fall off as $r^{-1}$. In other words, they co-exist, but one can define three regions: one where the near-field dominates (at small $r$), one where the far-field dominates (at large $r$) and a third where they are of comparable magnitude and must both be considered.

There is no black & white dividing line between these regions - it depends on the purpose of your calculations/investigations. Typically one says that the far-field radiation field dominates if $r \gg \lambda/2\pi$ (where $\lambda$ is the wavelength of the far-field radiation) and if $r \gg l$, where $l$ is the dimension of the antenna. What $\gg$ means is entirely a matter for you.

  • $\begingroup$ Why does the amplitude of the near field fall off as r-2, while that of the far field falls off at r-1 ? $\endgroup$ – John Petrovic Jan 16 '17 at 15:02
  • $\begingroup$ Because that is the definition of the "near field". If it fell of as $r^{-1}$ then it wouldn't be the near field @JohnPetrovic If you are asking why there is a near field and a far field, well think about the field of an accelerating charge - there will obviously be some static "Coulomb component" that varies as $r^{-2}$. $\endgroup$ – Rob Jeffries Jan 16 '17 at 16:07
  • $\begingroup$ Think about this. A single moving electron generates a magnetic field around it that moves (i.e. changes in space) with the motion of the electron. Many moving electrons in an antenna produce a magnetic field in the vicinity of the antenna. How does this lead to the formation of an independently traveling electromagnetic radio wave that no longer interacts with the moving electrons in the antenna? This is what I am trying to understand. $\endgroup$ – John Petrovic Jan 16 '17 at 16:22
  • $\begingroup$ Because a "moving" electron doesn't generate EM waves, only an accelerating electron does that. And if you are talking about a linear antenna then of course the far-field component is zero along the direction of the antenna. $\endgroup$ – Rob Jeffries Jan 16 '17 at 16:35
  • $\begingroup$ Tell me how an accelerating electron produces an electromagnetic wave, that's what I want to understand. $\endgroup$ – John Petrovic Jan 16 '17 at 17:06

I read a bit the wiki article on near and far field and antennas.

To start with an electromagnetic wave always travels with velocity c , no matter how the interference patterns of continuous emissions appear.

An analogy: When one is near a lamp one can see the emitting wires or diodes, which means that it is not a plane wave of light but it has a particular structure , at a distance the light source follows the 1/r^2 law of dispersion from a point source and can be described by a spherical wave.

I would like to know how far field traveling electromagnetic radio waves separate from the near field non-traveling electromagnetic fields at a radio transmitter antenna .

It is not a matter of separating electromagnetic waves, but of diminishing importance of the interference patterns that control the near fields due to the small distance from the source and the shape of the source, the wavelength being the gauge.


What you call "non traveling " wave is a misnomer. It is a non traveling diffraction pattern. Diffraction patterns are dependent on the continuous emission of the electromagnetic wave from the particular topology of the antenna. At a distance, as the beam opens in angle and the energy density drops these diffraction patterns disappear and a dipole pattern of energy density dominates , if the antenna is a dipole.

  • $\begingroup$ This diagram is from Wikipedia near and far field waves. But it does not answer my question. My question is what happens at the line in the diagram that separates fresnel radiative waves from far field waves. $\endgroup$ – John Petrovic Jan 15 '17 at 19:35
  • $\begingroup$ The diagram is for people who are not clear for far field and near field meanings. My answer states that there is NO line, rather the line is a few wavelengths wide, of continuously fading interference. The near field interferences fade away in a few wavelengths due to the opening rays and only the radial 1/r^2 behavior remains. Do not confuse interference patterns of em waves with the em waves. $\endgroup$ – anna v Jan 16 '17 at 4:20
  • $\begingroup$ Interference effects happen when multiple wave patterns interact with each other. Since one radio wave is generated by each cycle of the charges moving in the antenna, where do these additional wave patterns come from? $\endgroup$ – John Petrovic Jan 16 '17 at 15:07
  • $\begingroup$ In electromagnetism the waves do not interact. They overlap, it is in the quantum mechanical regime, the wavefunctions are superposed, and their comlex conjugate squared show the interference patterns in the energy content of the beam. The classical wave also, but it is not an interaction, it is an overlap because by their construction the waves are in phase . $\endgroup$ – anna v Jan 16 '17 at 15:47

A perfect oscillator may create electric field in a capacitor, and magnetic field in an inductor, and (being 'perfect') will not radiate. That's not useful for a radio, so we add a component, the antenna, that has both electric and magnetic fields, but which ALSO has a loss element, a resistive component to its impedance.

The energy loss in an antenna is intended to be radiation out into space of an electromagnetic wave. The OTHER parts of the energy provided to the antenna are NOT lost, they just cycle energy from inductor/magnetic field energy to capacitive/electric field energy, back and forth, just like an inductor and capacitor (LC tank circuit).

To evaluate an antenna design, the how-much-power-is-emitted question requires a calculation of the fields near the antenna, due to the supplied current at the transmitter frequency. That is important, because nearby objects (the transmitter tower, the ground) and the shape of the antenna elements, determines beam direction. The radiated intensity, though, in any given direction, will only depend on the way those local fields couple to a wave-equation traveling wave solution. So, after solving local fields, one also finds a second part solution, to the far field traveling wave that delivers radio to your receiver.

The calculation has two parts. The physical situation is only ONE part, but we can use approximations at a distance that wouldn't work up close to the antenna (where we care about different answers, anyhow).


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